Cohomology $H^1(G{L_n}({\mathbb{Z}}/{2\mathbb{Z}}),({\mathbb{Z}}/{2\mathbb{Z}})^n))$

Question

$\newcommand{\Z}{\mathbb{Z}}$ In my thesis I had a problem that could be solved by proving that $H^1(GL_n(\Z/2\Z),(\Z/2\Z)^n)$ is trivial for all $n\geq 2$. This is something my supervisor said, but I know nothing about cohomology of groups (I know a bit over sheaf cohomology though). I do not want to learn this new subject since I am almost finished and do not need it.

This result should be a well-known result according to my supervisor, however I could not find it in any books. Do you guys know some reference or a simple argument to solve this?

Answer 1

Here is a solution which is relatively simple considering the basics of group cohomology. You can look these basics up in the book Cohomology of groups by K. S. Brown.

I assume that $GL_n(\mathbb{F}_2)$ has the usual action over $\mathbb{F}_2^n$. Note that $GL_n(\mathbb{F}_2)$ acts transitively over the standard basis of $\mathbb{F}_2^n$ and the isotropy subgroup of the vector $(1,0,\ldots,0)$ is the subgroup $R_n$ of invertible matrices whose first column is $(1,0,\ldots,0)$. Therefore we have an isomorphism of $\mathbb{F}_2 GL_n(\mathbb{F}_2)$-modules:

$$ \mathbb{F}_2^n \cong \mathbb{F}_2 (GL_n(\mathbb{F}_2)/R_n) $$

and therefore

$$ H^1(GL_n(\mathbb{F}_2);\mathbb{F}_2^n) \cong H^1(GL_n(\mathbb{F}_2);\mathbb{F}_2 (GL_n(\mathbb{F}_2)/R_n)) \cong H^1(R_n;\mathbb{F}_2) $$

Note that any matrix in $R_n$ has the form (in blocks)

$$ \left( \begin{array}{cc} 1 & v \\ 0 & A \end{array} \right) $$

where $0$ is the zero column vector in $\mathbb{F}_2^{n-1}$, $A$ belongs to $GL_{n-1}(\mathbb{F}_2)$ and $v$ is a row vector in $\mathbb{F}_2^{n-1}$. Multiplication by blocks shows that $R_n$ is isomorphic to the semidirect product $\mathbb{F}_2^{n-1} \rtimes GL_{n-1}(\mathbb{F}_2)$ with the usual action of $GL_{n-1}(\mathbb{F}_2)$ over $\mathbb{F}_2^{n-1}$ on the right.

The extension of groups $ 1 \to \mathbb{F}_2^{n-1} \to R_n \to GL_{n-1}(\mathbb{F}_2) \to 1$ gives us a five-term exact sequence (only the first three are relevant for what follows)

$$ 0 \to H^1(GL_{n-1}(\mathbb{F}_2);\mathbb{F}_2) \to H^1(R_n;\mathbb{F}_2) \to H^1(\mathbb{F}_2^{n-1};\mathbb{F}_2)^{GL_{n-1}(\mathbb{F}_2)} \to \ldots $$

Now in general $H^1(G;\mathbb{F}_2) \cong \text{Hom}(G_{\text{ab}},\mathbb{Z}/2)$, where $G_{\text{ab}}$ is the abelianization of $G$. In particular,

$$ H^1(\mathbb{F}_2^{n-1};\mathbb{F}_2) \cong \text{Hom}(\mathbb{F}_2^{n-1},\mathbb{Z}/2) \cong \mathbb{F}_2^{n-1} $$

Under this isomorphism, the action of $GL_{n-1}(\mathbb{F}_2)$ becomes the usual action on $\mathbb{F}_2^{n-1}$ on the right. So our exact sequence becomes

$$ 0 \to \text{Hom}(GL_{n-1}(\mathbb{F}_2)_{\text{ab}},\mathbb{Z}/2) \to H^1(R_n;\mathbb{F}_2) \to (\mathbb{F}_2^{n-1})^{GL_{n-1}(\mathbb{F}_2)} $$

And note that $(\mathbb{F}_2^{n-1})^{GL_{n-1}(\mathbb{F}_2)}$ is zero if $n \geq 3$.

Now we have different cases:

• $R_2 \cong \mathbb{Z}/2$ and so we get

$$ H^1(GL_2(\mathbb{F}_2);\mathbb{F}_2^2) \cong H^1(R_2;\mathbb{F}_2) \cong \mathbb{F}_2 $$

• $GL_2(\mathbb{F}_2) \cong \Sigma_3$, whose abelianization is $\mathbb{Z}/2$ and so

$$ H^1(GL_3(\mathbb{F}_2);\mathbb{F}_2^3) \cong H^1(R_3;\mathbb{F}_3) \cong \text{Hom}(\mathbb{Z}/2,\mathbb{F}_2) \cong \mathbb{F}_2 $$

• $GL_k(\mathbb{F}_2)_{\text{ab}}$ is trivial if $k>2$, so if $ n >3$ then

$$ H^1(GL_n(\mathbb{F}_2),\mathbb{F}_2^n) \cong H^1(R_n;\mathbb{F}_2) = 0 $$